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Ta có:
\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)
\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+16+8\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+16\right)+8\left(x^2+10x+16\right)+16\)
\(=\left(x^2+10x+16\right)^2+2.\left(x^2+10x+16\right).4+4^2\)
\(=\left(x^2+10x+16+4\right)^2=\left(x^2+10+20\right)^2\)
k nha!!
\(\text{( x + 2 ) ( x + 4 ) ( x + 6 ) ( x + 8 ) + 16}\)
\(\text{Phân tích thành nhân tử :}\)
\(\left(x^2+10x+20\right)^2\)
(x+2)(x+4)(x+6)(x+8)+16
=(x+2)(x+8)(x+4)(x+6)+16
=(x2+10x+16)(x2+10x+24)+16
đặt t=x2+10x+16 ta được:
t.(t+8)+16
=t2+8t+16
=(t+4)2
thay t=x2+10x+16 ta được:
(x2+10x+16)2
=[(x+2)(x+8)]2
=(x+2)2(x+8)2
vậy (x+2)(x+4)(x+6)(x+8)+16 =(x+2)2(x+8)2
(x+2)(x+4)(x+6)(x+8)+16
=(x+2)(x+8)(x+4)(x+6)+16
=(x2+10x+16)(x2+10x+24)+16
đặt t=x2+10x+16 ta được:
t.(t+8)+16
=t2+8t+16
=(t+4)2
thay t=x2+10x+16 ta được:
(x2+10x+16)2
=[(x+2)(x+8)]2
=(x+2)2(x+8)2
vậy (x+2)(x+4)(x+6)(x+8)+16 =(x+2)2(x+8)2
x^40+2.x^20+9 = [x^20 +3]^2 - 4x^20 = [x^20+3]^2 -[2x^10]^2 = [x^20-2x^10+3].[x^20+2x^10+3]
x^12+x^6+1 = x^12 + 2x^6 +1 - x^6 = [x^6 +1]^2 -[x^3]^2 = [x^6 -x^3 +1].[x^6+x^3+1]
x^16+x^8+1 =[x^8+1]^2 - [x^4]^2 = [x^8-x^4+1].[x^8+x^4+1]
x^4+x^2+1 = x^4+2x^2+1 - x^2 = [x^2+1]^2-x^2 = [x^2-x+1].[x^2+x+1]
Đề Phân tích đa thức thành nhân tử 1/(1 - x )+ 1/(1+x)+2/(1+x^2)+ 4/(1+x^4)+8/(1+x^8) - 16/(1+ x^16)
Ta có: (x+2)(x+4)(x+6)(x+8)+16
=[(x+2)(x+8)]+[(x+4)(x+6)]+16
\(=\left[x^2+10x+16\right]\left[x^2+10x+24\right]+16\) (1)
Đặt \(x^2+10x+16=t\), khi đó (1) trở thành:
\(t\left(t+8\right)+16=t^2+8t+16=\left(t+4\right)^2\)
Thay \(x^2+10x+16=t\), ta có: \(\left(x^2+10x+16+4\right)^2=\left(x^2+10x+20\right)^2\)
Có gì đó sai sai á nhờ :vv?
( x + 2 )( x + 4 )( x + 6 )( x + 8 ) + 16
= [ ( x + 2 )( x + 8 ) ][ ( x + 4 )( x + 6 ) ] + 16
= ( x2 + 10x + 16 )( x2 + 10x + 24 ) + 16 (*)
Đặt t = x2 + 10x + 20
(*) <=> ( t - 4 )( t + 4 ) + 16
= t2 - 16 + 16
= t2 = ( x2 + 10x + 20 )2
x 16 + x 8 − 2 = ( x 8 ) 2 + x 8 − 2 = ( x 8 − 1 ) ( x 8 + 2 ) = ( x 4 − 1 ) ( x 4 + 1 ) ( x 8 + 2 ) = ( x 2 − 1 ) ( x 2 + 1 ) ( x 4 + 1 ) ( x 8 + 2 ) = ( x − 1 ) ( x + 1 ) ( x 2 + 1 ) ( x 4 + 1 ) ( x 8 + 2 )
Ta có : (x+2)(x+4)(x+6)(x+8) + 16
=[(x+2).(x+8)].[(x+4)(x+6)]+16
=(x2+10x+16).(x2+10x+24)+16 (1)
Đặt x^2+10x+16=a thì (1) trở thành:
a.(a+8)+16=a2+8a+16=(a+4)2=(x^2+10x+20)2
a, Cách 1 : \(x^2+5x+6=x^2+2x+3x+6=\left(x+2\right)\left(x+3\right)\)
Cách 2 : \(x^2+5x+6=x^2+2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}+6\)
\(=\left(x+\frac{5}{2}\right)^2-\frac{1}{4}=\left(x+2\right)\left(x+3\right)\)
b, Cách 1 : \(x^2-x-6=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\)
Cách 2 : \(x^2-x-6=x^2-x+\frac{1}{4}-\frac{1}{4}-6=\left(x-\frac{1}{2}\right)^2-\frac{25}{4}=\left(x-3\right)\left(x+2\right)\)
c, Cách 1 : \(x^2+6x+8=x^2+4x+2x+8=\left(x+2\right)\left(x+4\right)\)
Cách 2 : \(x^2+6x+8=x^2+6x+9-1=\left(x+3\right)^2-1=\left(x+2\right)\left(x+4\right)\)
d, Cách 1 : \(x^2-2x-8=x^2+2x-4x-8=\left(x-4\right)\left(x+2\right)\)
Cách 2 : \(x^2-2x-8=x^2-2x+1-9=\left(x-1\right)^2-9=\left(x-4\right)\left(x+2\right)\)